PHY 2311

PHYSICS 231Lecture 8: Forces, forces & examples

Remco ZegersWalk-in hour: Monday 11:30-13:30

Helproom

PHY 2312

Newton’s Laws

First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity.

Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma

If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F12=-F21

PHY 2313

Forces seen in the previous lecture

• Gravity: Force between massive objects• Normal force: Elasticity force from supporting surface

Fg//=mgsinFgL=mgcos

Fg=mg

n=-FgL

PHY 2314

Gravity, mass and weights.

Weight=mass times gravitational accelerationFg(N)=M(kg) g(m/s2)

Newton’s law of universal gravitation:Fgravitation=Gm1m2/r2

G=6.67·10-11 Nm2/kg2

For objects on the surface of the earth:•m1=mearth=fixed

•r=“radius” of earth=fixed•The earth is not a point object relative to m2

PHY 2315

Measuring mass and weight.

Given that gearth=9.81 m/s2, gsun=274 m/s2, gmoon=1.67 m/s2,what is the mass of a person on the sun and moon if hismass on earth is 70 kg? And what is his weight on each ofthe three surfaces?

• The mass is the same on each of the surfaces

• On Earth: w=686.7 N• On the Moon: w=116.7 N• On the Sun: 19180 N

PHY 2316

Jumping!

The pelvis has a mass of 30.0 kg. Whatis its acceleration?

Decompose all forces in x and y directions

Force x (N) y (N)300 N -122 -274690 N 236 648Weight 0 -249Resultant 114 N 80.3 N

Total Force: F=(1142+80.32)=139 NDirection: =tan-1(Fy/Fx)=35.2o

Acceleration: a=F/m=139/30.0=4.65 m/s2

PHY 2317

TensionT The magnitude of the force T

acting on the crate, is the sameas the tension in the rope.

Spring-scale

You could measure the tension by insertinga spring-scale...

PHY 2318

Newton’s second law and tensionm1

m2

No friction.n

Fg

T

T

Fg

Object 1: F=m1a, so T=m1aObject 2: F=m2a, so Fg-T=m2a

m2g-T=m2aCombine 1&2 (Tension is the same): a=m2g/(m1+m2)

What is the acceleration ofthe objects?

PHY 2319

Problem

T

Draw the forces: what is positive &negative???

Fg

Fg

T

For 3.00 kg mass: F=ma T-9.813.00=3.00a

For 5.00 kg mass: F=ma 9.815.00-T=5.00a

What is the tension in the string andwhat will be the acceleration of thetwo masses?

T=36.8 Na=2.45 m/s2

PHY 23110

Friction

Friction are the forces acting on an object due to interaction

with the surroundings (air-friction, ground-friction etc).Two variants:•Static Friction: as long as an external force (F) trying to make an object move is smaller than fs,max, the static friction fs equals F but is pointing in the opposite direction: no movement!

fs,max=sn s=coefficient of static friction

•Kinetic Friction: After F has surpassed fs,max, the object starts moving but there is still friction. However, the friction will be less than fs,max!

fk=kn k=coefficient of kinetic friction

PHY 23111

PHY 23112

Problem

Fg//=mgsinFgL=mgcos

Fg=mg

n=-FgL

Fs,kA)If s=1.0, what isthe angle for which theblock just starts to slide?B)The block starts moving. Given that k=0.5, what is theacceleration of the block?

A) Parallel direction: mgsin-sn=0 (F=ma) Perpendicular direction: mgcos-n=0 so n=mgcos Combine: mgsin-smgcos=0 s=sin/cos=tan=1 so =45oB) Parallel direction: mgsin(45o)- smgcos(45o)=ma (F=ma) g(½2-¼2)=a so a=g¼2

PHY 23113

All the forces come together...

Fg

Tn

Fk

Fg

T

If a=3.30 m/s2 (the 12kg blockis moving downward), what isthe value of k?

For the 7 kg block parallel to the slope:T-mgsin-kmgcos=maFor the 12 kg block: Mg-T=Ma

25.0cos

sin)(

mg

mamgagMkSolve for k

PHY 23114

General strategy

•If not given, make a drawing of the problem.•Put all the relevant forces in the drawing, object by object.

• Think about the axis• Think about the signs

•Decompose the forces in direction parallel to the motion and perpendicular to it.

•Write down Newton’s first law for forces in the parallel direction and perpendicular direction.

•Solve for the unknowns.•Check whether your answer makes sense.

PHY 23115

Next Lecture:

Revision and go through an exam

May the Force be with you!